Compute the multiplicative inverse of in
where .
Because , we have . It is easy to see that , so is the inverse.
However, it is not always so easy to solve for inverses by inspection. The surefire way to do this is to use linear algebra, as I will demonstrate.
You want (we only go up to the second degree since )
Expanding the left hand side gives
So we want to find such that . Although it is obvious what the solution is in this case, I'll write down the general way to do it:
With respect to the basis , this is equivalent to solving the linear system
Which has solutions . Hence the inverse is , which is the same answer as we got from inspection.
How do you guys post equations so neatly?
By copying and pasting the original post, this is what I get:
"Compute the multiplicative inverse of a(x) = \overline{x^2 + x + 1} in \mathbb{Z}_5/<p(x)>
where p(x) = x^3 + x^2 + x + 1 = (x^2 + 1)(x +1) . "
Hi HowDoIMath,
You need to use the Latex language for Maths expressions - see Latex Help and download the very helpful tutorials or help sheets.
Note that you need [TEX] and then the same with a / before the T around the math expressions to get Latex to work
Peter